【leetcode】Combination Sum II

Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

思路：
由于要去重，因此，下一次的递归的点为i+1，同时还得注意，一样的组合不能重复出现，因此需要下式来去重：

while(i<candidates.size()-1 && candidates[i]==candidates[i+1]) i++; 

代码：

class Solution {public:    vector<vector<int> > res;    vector<int> temp;    void combination(vector<int>& candidates, int num, int tempsum, int target)    {        if(tempsum>target) return;        if(tempsum==target)        {            res.push_back(temp);            return;        }        for(int i=num; i<candidates.size();i++)        {            tempsum+=candidates[i];            temp.push_back(candidates[i]);            combination(candidates,i+1,tempsum,target);            temp.pop_back();            tempsum-=candidates[i];            while(i<candidates.size()-1 && candidates[i]==candidates[i+1])                i++;        }    }    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        if(candidates.size()==0) return res;        sort(candidates.begin(),candidates.end());        combination(candidates, 0, 0,target);        return res;    }};